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The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be
A
$\sqrt{2}T$
B
T
C
$\frac{T}{\sqrt{2}}$
D
2T
Detailed Solution
Time period of a spring–mass system: $T = 2\pi\sqrt{\frac{M}{k}}$, so $T \propto \sqrt{M}$
When another mass M is added, the total mass becomes 2M.
$\frac{T_2}{T_1} = \sqrt{\frac{M + M}{M}} = \sqrt{2}$
$T_2 = \sqrt{2}T$
When another mass M is added, the total mass becomes 2M.
$\frac{T_2}{T_1} = \sqrt{\frac{M + M}{M}} = \sqrt{2}$
$T_2 = \sqrt{2}T$